Chapter 31, Problem 1Q

Figure 31-19 shows three oscillating LC circuits with identical inductors and capacitors At a particular time, the charges on the capacitor plates (and thus the electric fields between the plates) are all at their maximum values Rank the circuits according to the time taken to fully discharge the capacitors during the oscillations, greatest first.

Figure 31-19 Question 1.

To determine

To find:

The rank of the circuits according to the time taken to fully discharge the capacitors during the oscillations.

Answer to Problem 1Q

Solution:

The rank of the circuits according to time taken to fully discharge the capacitors during the oscillations is circuit b, circuit a, circuit c.

Explanation of Solution

1) Concept:

The charging and discharging of a capacitor in a LC circuit is like an oscillatory motion. The period of these oscillations depends upon the values of the inductance and the capacitance in the circuit.

2) Formula:

i) $\omega = \frac{2\pi }{T}$

ii) $\omega = \frac{1}{\sqrt{LC}}$

3) Given:

i) The inductors and capacitors in the three circuits are identical.

ii) The two capacitors in the circuit b are in parallel combination.

iii) The two capacitors in the circuit c are in series combination.

4) Calculations:

a) Consider circuit b. The two capacitors are connected in parallel combination. Hence the effective capacitance of the circuit is

$C= C_1+C_2$

Since both the capacitors are identical, the effective capacitance is

$C_b= C_1+C_2=2C$

b) Now, consider circuit c. The two capacitors are connected in series combination. Hence the effective capacitance of the circuit is

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

Since both the capacitors are identical, the effective capacitance is

$\frac{1}{C_c}=\frac{1}{C_1}+\frac{1}{C_2}$

$\therefore \frac{1}{C_c}=\frac{C_1+C_2}{C_1{C}_2}=\frac{2C}{C^2}=\frac{2}{C}$

$\therefore C_c=\frac{C}{2}$

c) The period of oscillations is calculated using the equation

$\omega = \frac{2\pi }{T}$

and $\omega = \frac{1}{\sqrt{LC}}$

i.e.,  $T= \frac{2\pi }{\omega }=2\pi \sqrt{LC}$

Thus, we see that  $T \propto \sqrt{LC}$

But since the inductors in the three circuits are identical, $T \propto \sqrt{C}$

Now, for circuit b, the effective capacitance is greatest among the three. Hence its period is also the greatest. Thus, time for the capacitor to discharge fully, which is $\frac{T}{4}$, will also be the greatest among the three.

For circuit a, the capacitance is C, which is smaller than that for circuit b. Hence the time for the discharge will also be smaller.

For circuit c, the effective capacitance is the smallest among the three. Hence the time required for complete discharge will also be the smallest.

Thus the ranks for the circuits are circuit b, circuit a, and then circuit c.

Conclusion:

The time required for the capacitor to discharge fully is $\frac{T}{4}$. The period of the oscillation is directly proportional to the square root of the effective capacitance of the circuit. This helps us determine the ranking of the circuits.